Question:A solution of AgNO3 (45 mL/0.45 M) was mixed with solution of NaCl (85 mL/1.35 x 10-2 M)
a) Calculate the ion product of the potential precipitate.
My answer:
mol AgNO3 = M x L
= 0.45 M x 0.045 M
= 0.02025
mol NaCl = M x L
= 0.0135 M x 0.085 L
= 0.0011475
(Ag^+) = mols / total volume
= 0.02025 / 0.130 L
= 0.1558
total volume = 85 mL + 45 mL
= 130 mL
= 0.130 L
(Cl^-) = mols / total volume
= 0.0011475 / 0.130 L
= 0.00883
Ion product = [Ag^+][Cl^-]
= [0.1558][0.00883]
= 0.001375
Is my answer correct?
It looks ok to me. One point I want to make here. Technically, the Ksp of a salt, in this case, AgCl, cannot be exceeded (unless a supersaturated solution is formed). That is, (Ag^+)(Cl^-) can not be larger than Ksp which is about 1.8 x 10^-10. That's why what you have done is called the ion product. Since the ion product exceeds Ksp, you know a ppt will occur. How much AgCl will ppt. Enough AgCl will come out of solution so that the final (Ag^+) x the final (Cl^-) will be equal to 1.8 x 10^-10. In a saturated solution, (Ag^+) = (Cl^-) = 1.34 x 10^-5 molar so most will ppt. The point I want to make is that the ion product thing is an artificial one in that Ksp can't be exceeded. But the ion product is a way of comparing Ksp and knowing if a ppt will occur.
Thanks for the tip
Yes, your answer is correct. The ion product of the potential precipitate is indeed 0.001375.
To calculate the ion product, you need to find the concentrations of the respective ions involved in the precipitation reaction. In this case, the ions involved are Ag+ (from AgNO3) and Cl- (from NaCl).
First, you calculated the moles of AgNO3 by multiplying the molarity (0.45 M) by the volume (45 mL converted to L). This gave you 0.02025 moles.
Next, you calculated the moles of NaCl by multiplying the molarity (1.35 x 10-2 M) by the volume (85 mL converted to L). This gave you 0.0011475 moles.
Then, you divided the moles of Ag+ by the total volume (130 mL converted to L) to get the concentration of Ag+, which is 0.1558 M.
Similarly, you divided the moles of Cl- by the total volume (130 mL converted to L) to get the concentration of Cl-, which is 0.00883 M.
Finally, you multiplied the concentration of Ag+ by the concentration of Cl- to get the ion product. This gave you 0.001375.
Therefore, the ion product of the potential precipitate is 0.001375.